Get all records from collection

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9

That simple question should have a lot of answers.
Strapi does not recommand to do that - it might be heavy if you have a lot of records.
Anyway, a lot of users do need it.
Here’s my solution. In this example, the content type is area.

I override the find function in the controller.
If the query has its parameter pagination.page=-1, it’s a flag that we want ALL the records.
This solution is handly because you don’t need to create an extra route (eg. api/area/all), but you could if you want: the main idea is that we have a new service method findAll.

src/api/area/services/area.ts :


import { fetchAllEntries } from '../../../utils/fetchAllEntries'

export default factories.createCoreService('api::area.area', ({strapi}) => ({

  async find(params) {
    //get ALL items
    if (params?.pagination?.page === '-1') {
      return this.findAll(params);
    }

    // Calling the default core controller
    return await super.find(params);

  },
  async findAll(params) {
    return fetchAllEntries('api::area.area',params);
  },

Rather than overriding the find method in the service, I would prefer use a middleware for this, but well, I haven’t find a way to achieve it.

src/utils/fetchAllEntries.ts:

export async function fetchAllEntries(type, query:any = {}) {
  const maxLimit = strapi.config.get('api.rest.maxLimit', 25);
  let allItems = [];
  let iterations = 0;
  let totalPages = 1;
  let pagination = {} as any;

  //If pagination was set
  delete query.pagination;

  do {

    // Fetch entries with pagination
    const response = await strapi.service(type).find({
      ...query,
      pagination:{
        page:iterations+1,
        pageSize:maxLimit
      }
    });

    allItems = [...allItems, ...response.results];

    if (response.pagination){
      pagination = {...response.pagination};
      totalPages = pagination.pageCount;
    }

    console.info(`fetchAllEntries ${type}... ${iterations+1}/${pagination.pageCount}`);

    iterations++;

  } while (iterations < totalPages);

  delete pagination.page;//we'll return all results so page is not a relevant info

  return {
    results:allItems,
    pagination:pagination
  };
}